What is the extreme f(t)=2e^{ln(x)}(t) ?

The extreme f(t)=2e^{ln(x)}(t) is Saddle(0,0)

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extreme f(t)=2e^{ln(x)}(t)

extreme points

f (t) = 2 e^{l n (x)} (t)

Saddle (0, 0)

Solution steps

Find the critical points:

(0, 0)

\frac{\partial ^{2} f}{\partial t ^{2}} = 0

D (t, x) = - 4

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Saddle (0, 0)

f (x) = \frac{2 ( x + 6 ) ^{2}}{x + 8}

f (x) = \frac{x ^{3} + x - 2}{x ^{2} - 4 x + 3}

\frac{5 x - 15}{2 x - 9}

f (x, y) = 12 x y - x^{2} - 3 y^{2}

f (x) = x^{3} - 12 x^{2} + 36 x - 3, 0 \leq x \leq 9

What is the extreme f(t)=2e^{ln(x)}(t) ?
The extreme f(t)=2e^{ln(x)}(t) is Saddle(0,0)